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QJEEM24Ma5b8a: JEE Main 2024 Mathematics - Class 11 JEEM24Ma5b8a Solution: If the coefficients of and in the expansion of are in the…

If the coefficients of x4,x5x^4, x^5 and x6x^6 in the expansion of (1+x)n(1+x)^n are in the arithmetic progression, then the maximum value of nn is:

Multiple Choice Options

  • A 28
  • B 21
  • C 7
  • D 14

Step-by-Step Verified Solution

amp;(1+x)n=nC0+nC1x1+nC2x2+…nCnxnamp;nC4,nC5&nC6 are in A.P. amp;nC5−nC4=nC6−nC5amp;⇒n!5!(n−5)!−n!4!(n−4)!=n!6!(n−6)!−n!5!(n−5)!amp;⇒30(n−9)(n−6)=5(n−4)(n−11)amp;⇒30n2−450n+1620=5n2amp;⇒1n−5[n−4−55(n−4)]=15[n−5−66(n−5)]amp;⇒n−95(n−4)=15[n−116]amp;⇒n2−21n+98=0amp;nmax⁡=14\begin{aligned}& (1+x)^n={ }^n C_0+{ }^n C_1 x^1+{ }^n C_2 x^2+\ldots{ }^n C_n x^n \\& { }^n C_4,{ }^n C_5 \&{ }^n C_6 \text { are in A.P. } \\& { }^n C_5-{ }^n C_4={ }^n C_6-{ }^n C_5 \\& \Rightarrow \frac{n!}{5!(n-5)!}-\frac{n!}{4!(n-4)!}=\frac{n!}{6!(n-6)!}-\frac{n!}{5!(n-5)!} \\& \Rightarrow 30(n-9)(n-6)=5(n-4)(n-11) \\& \Rightarrow 30 n^2-450 n+1620=5 n^2 \\& \Rightarrow \frac{1}{n-5}\left[\frac{n-4-5}{5(n-4)}\right]=\frac{1}{5}\left[\frac{n-5-6}{6(n-5)}\right] \\& \Rightarrow \frac{n-9}{5(n-4)}=\frac{1}{5}\left[\frac{n-11}{6}\right] \\& \Rightarrow n^2-21 n+98=0 \\& n_{\max }=14\end{aligned}

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